In the circuit shown below,the maximum Zener | vedclass

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(A) The Zener diode current $I_Z$ is given by $I_Z = I - I_L$,where $I$ is the total current from the source and $I_L$ is the load current.To maximize

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$9$

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(A) The Zener diode current $I_Z$ is given by $I_Z = I - I_L$,where $I$ is the total current from the source and $I_L$ is the load current.To maximize $I_Z$,we must maximize the total current $I$. Since $I = \frac{V_{in} - V_Z}{R}$,$I$ is maximum when $V_{in}$ is maximum $(V_{in} = 120 	ext{ V})$.$I_{max} = \frac{120 	ext{ V} - 60 	ext{ V}}{4000 \ \Omega} = \frac{60 	ext{ V}}{4000 \ \Omega} = 0.015 	ext{ A} = 15 	ext{ mA}$.The load current $I_L$ is constant because the voltage across the load resistor is fixed at $V_Z = 60 	ext{ V}$.$I_L = \frac{V_Z}{R_L} = \frac{60 	ext{ V}}{10000 \ \Omega} = 0.006 	ext{ A} = 6 	ext{ mA}$.Therefore,the maximum Zener diode current is $I_{Z,max} = I_{max} - I_L = 15 	ext{ mA} - 6 	ext{ mA} = 9 	ext{ mA}$.

In the circuit shown below,the maximum Zener diode current will be $..... \text{mA}$.

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